To start with, we write the fractional part condition as follows:
\([\alpha p^\theta - \delta _1] - [\alpha p^\theta - \delta _2] &=&{\left\lbrace \begin{array}{ll} 1 & \mbox{if $\delta _1 \le \lbrace  \alpha p^\theta \rbrace  < \delta _2$;} \\0 & \mbox{otherwise.} \end{array}\right.}\) 
First, we obtain the result when \(x_j<p\le x_{j+1}\)  for
\(x_j:= [x(1+j/B)]+1/2\)  for \(j=0,\ldots , B\)  and \(B=[\omega ]\) . Summing over \(j\)  then establishes the result for \(x<p\le 2x\) .
With \(\delta = \delta _2 - \delta _1\) , the length of the interval, we set
\(U_{-} &:=& \frac{\alpha x_j^\theta }{\delta }, \qquad U_{+} := \frac{\alpha x_{j+1}^\theta }{\delta }.\) 
Then for \(x_j< p \le x_{j+1}\) , we have
\(\alpha p^{\theta } \left(1- \frac{1}{U_{-}} \right) - \delta _1 \le \alpha p^{\theta } - \delta _2 \le \alpha p^{\theta } \left(1- \frac{1}{U_{+}} \right) - \delta _1.\) 
Note that we are interested in the sum
\(S:=\sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}} [\alpha p^{\theta } - \delta _1 ] - [\alpha p^{\theta } - \delta _2]\) 
in order to bound the number of primes \(x_j <p\le x_{j+1}\)  such that \(\sigma _p\in C\)  and \(\delta _1 \le \lbrace \alpha p^\theta \rbrace  < \delta _2.\)  This implies
\(S &\ge & \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}} \left[\alpha p^\theta - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{+}} \right) - \delta _1 \right]\) 
and
\(S &\le & \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}} \left[\alpha p^\theta - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{-}} \right) - \delta _1 \right].\) 
Therefore, using
\( \frac{\alpha p^\theta }{U_{\pm }}=\delta +\operatorname{O}\left(\frac{\delta }{\omega }\right),\) 
in order to obtain the claimed asymptotics for \(S\) , it suffices to prove
\(&\nonumber \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}} \left[\alpha p^\theta - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) - \delta _1 \right]-\frac{\alpha p ^\theta }{U_{\pm }} \\&\nonumber \ll \frac{|C|}{|G|}\log x \left( (\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{(3+\theta )/4} + \frac{\delta n_L}{\alpha ^{1/2}}x^{1-\theta /2}\log x\right)\\& \quad + (\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{(1+\theta )/4}\log x.\) 

We may write the summand
\(\left[\alpha p^\theta - \delta _1 \right] - \left[ \alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) - \delta _1 \right] &= \sum _{m \le \alpha p^\theta -\delta _1}1\ -\sum _{m \le \alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right)-\delta _1 } 1.\) 

Define the sequence \((a_m)_{m\in \mathbb {N}}\)
\(a_m:= {\left\lbrace \begin{array}{ll}1 &\text{ if } \frac{1}{3} \alpha x^\theta -\delta _1 < m \le 3 \alpha x^\theta -\delta _1\\0 &\text{ otherwise}.\end{array}\right.}\) 
Set
\(A_{\delta _1}(M) := \sum _{m \le M-\delta _1} a_m = \sum _{m\ge 1}a_m \ f\left( \frac{m+\delta _1}{M}\right)\) 
where
\(f(y)= {\left\lbrace \begin{array}{ll}1 &\text{ if } 0<y<1\\1/2 & \text{ if } y=1\\0 &\text{ if } y>1\end{array}\right.}\) 
Using the inverse Mellin transform, for \(\sigma >0\) , we write
\(f(y) = \frac{1}{2\pi i}\int _{\sigma -i\infty }^{\sigma + i\infty } \frac{y^{-s}}{s} ~ds.\) 
For \(y= (m+\delta _1)/M\) , this gives
\(A_{\delta _1}(M) = \frac{1}{2\pi i}\int _{\sigma -i\infty }^{\sigma + i\infty } \sum _{m\ge 1} \frac{a_m}{(m+\delta _1)^s}\frac{M^s}{s}~dx.\) 
Thus, using the definition of the sequence \((a_m)\) , we rewrite (REF ) as
\(\left[\alpha p^\theta - \delta _1 \right] - \left[ \alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) - \delta _1 \right]&=\sum _{m \le \alpha p^\theta -\delta _1 }a_m\ - \sum _{m \le (\alpha p^\theta ) \left(1-\frac{1}{U_{\pm }} \right)-\delta _1 }a_m\\&= A_{\delta _1}(\alpha p^{\theta }) - A_{\delta _1}\left(\alpha p^{\theta }\left(1-\frac{1}{U_{\pm }} \right) \right).\) 

In order to estimate the integrals \(A_{\delta _1}(\alpha p^{\theta })\)  and \(A_{\delta _1}\left(\alpha p^{\theta }\left(1-\frac{1}{U_{\pm }} \right) \right)\) , given by (REF ), we make use of the truncated Perron's formula [1]}, and obtain
\(&\nonumber \left[\alpha p^\theta - \delta _1 \right] - \left[ (\alpha p^\theta ) \left(1-\frac{1}{U_{\pm }} \right) - \delta _1 \right]= \frac{1}{2\pi i} \int _{1/2-i T_1}^{1/2+i T_1} L(s) H(s) \ p^{\theta s} ~ds \\& \quad +\operatorname{O}\Bigg (\sum _{\frac{1}{3} \alpha x^\theta < m+\delta _1 \le 3 \alpha x^\theta }\! \min \left\lbrace 1,T_1^{-1}\left|\log \frac{\alpha p^\theta }{m+\delta _1}\right|^{-1}\right\rbrace \Bigg ) \\& \quad +\operatorname{O}\Bigg (\sum _{\frac{1}{3} \alpha x^\theta < m+\delta _1 \le 3 \alpha x^\theta }\!\min \left\lbrace 1,T_1^{-1}\left|\log \frac{\alpha p^\theta }{m+\delta _1} \left(1-\frac{1}{U_{\pm }}\right)\right|^{-1} \right\rbrace \Bigg )\\&= \frac{1}{2\pi i} \int _{1/2-i T_1}^{1/2+i T_1} L(s) H(s) p^{\theta s} ~ds +\operatorname{O}\Bigg (\frac{\alpha x^{\theta } }{T_1}\log (\alpha x^{\theta })\Bigg ),\\\) 

where
\(H(s) := \frac{1}{s}\left(1-\left(1-\frac{1}{U_{\pm }}\right)^s\right) \ll \frac{1}{U_{\pm }}\) 
and
\(L(s) := \alpha ^s \sum _{\frac{1}{3} \alpha x^\theta - \delta _1 < m \le 3 \alpha x^\theta - \delta _1}\frac{1}{ (m+\delta _1)^{s}}.\) 
Summing over the primes, we find
\(&\nonumber \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}}\Bigg (\left[\alpha p^\theta \ - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) -\delta _1 \right]\Bigg ) \\&\hspace{18.06749pt} = \frac{1}{2\pi i} \int _{1/2-iT_1}^{1/2+iT_1} L(s) H(s) F(-\theta s)~ds + \operatorname{O}\left( \frac{\alpha x^{\theta }}{T_1} \log (\alpha x^{\theta })\pi _C(x_j,L) \right)\) 

where
\( F(s) := \sum _{\begin{array}{c}x_j<p\le x_{j+1} \\ \sigma _p \in C\end{array}} p^{- s} \qquad \text{ and } \qquad \pi _C(x_j,L) := \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}}1.\) 
First, we compute the above integral in the smaller range, up to \(T_0 :=\alpha x^\theta \) . Observe that in the range \(|t|\le T_0\) , we have
\(H(s) =\frac{1}{U_{\pm }} + \operatorname{O}\left(\frac{|s-1|}{U_{\pm }^2} \right) \) 
and
\( L(s) = \alpha ^s \frac{(3 \alpha x^\theta )^{1-s} - (\alpha x^\theta /3)^{1-s}}{1-s} + \operatorname{O}\left( x^{-\theta \Re (s)} \right)\) 
Therefore, for the integral in (REF ) in the range \(|t|\le T_0\) , we have
\(& \nonumber \frac{1}{2\pi i} \int _{1/2-iT_0}^{1/2+iT_0} L(s) H(s)F(-\theta s)\,ds \\ &= \frac{1}{U_{\pm }} \sum _{\begin{array}{c}x_j < p \le x_{j+1} \\ \sigma _p \in C\end{array}} \alpha p^{\theta } \; \frac{1}{2\pi i} \int _{1/2-iT_0}^{1/2+iT_0} (\alpha p^{\theta })^{s-1} \frac{(3 \alpha x^\theta )^{1-s} - (\alpha x^\theta /3)^{1-s}}{1-s}~ds \\&\quad + \operatorname{O}\left( \frac{\delta }{\alpha } x^{-3\theta /2} \int _{-T_0}^{T_0} |F(-\theta /2 - i \theta t)| \; dt \right).\) 

Using Cauchy-Schwarz inequality, the error term () is bounded by
\( &&\ll \frac{\delta }{\alpha } x^{-3\theta /2} T_0^{1/2} \left( \int _{-T_0}^{T_0} |F(-\theta /2 - i \theta t)|^2~dt \right)^{1/2}.\) 
Also, by applying the mean value theorem for Dirichlet polynomials, we have
\(\int _{-T_0}^{T_0} \left|F(-\theta /2-i \theta t) \right|^2 ~dt \ll x^{\theta }\pi _C(x_j,L) \;\left( T_0 + x/ \omega \right).\) 
Inserting this estimate in (REF ), the error term () is bounded by
\(\nonumber &\ll \delta T_0^{1/2}(\alpha x^\theta )^{-1} \pi _C(x_j,L)^{1/2}\left(T_0^{1/2}+x^{1/2}/\omega ^{1/2}\right)\\&\ll \delta \pi _C(x_j,L)^{1/2}\left(1+x^{(1-\theta )/2}(\alpha \omega )^{-1/2}\right),\) 

using \(T_0=\alpha x^\theta \) . We now compute the integral in (REF ) to obtain the desired main term. The change of variable \(w=1-s\)  yields
\(\frac{1}{U_{\pm }} \sum _{\begin{array}{c}x_j< p \le x_{j+1}\\ \sigma _p \in C\end{array}} \alpha p^{\theta } \frac{1}{2\pi i} \int _{1/2-iT_0}^{1/2+iT_0} \frac{1}{w} \left(\left(\frac{3 x^\theta }{ p^\theta }\right)^{w} - \left(\frac{ x^\theta }{3p^\theta }\right)^{w} \right)~dw.\) 

By Perron's formula, for all values \(x_j < p \le x_{j+1}\) ,
\(\frac{1}{2\pi i} \int _{1/2-iT_0}^{1/2+iT_0} \frac{1}{w} \left(\left(\frac{3 x^\theta }{ p^\theta }\right)^{w} - \left(\frac{ x^\theta }{3p^\theta }\right)^{w} \right)~dw =1 + \operatorname{O}\left( \frac{1}{T_0} \right).\) 
Therefore (REF ) becomes
\(\sum _{\begin{array}{c}x_j < p \le x_{j+1}\\ \sigma _p \in C\end{array}} \frac{\alpha p^{\theta }}{U_{\pm }} \ + \operatorname{O}\left( \frac{1}{T_0 U_{\pm }} \sum _{\begin{array}{c}x_j < p \le x_{j+1}\\ \sigma _p \in C\end{array}} \alpha p^{\theta } \right).\) 
Using \(T_0 =\alpha x^{\theta }\)  and collecting the terms from (REF ), (REF ) and (REF ), we obtain
\(&\nonumber \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}} \left( \left[\alpha p^\theta \ - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) -\delta _1 \right] - \frac{\alpha p^{\theta }}{U_{\pm }}\right) \\&\nonumber \ll \delta \pi _C(x_j,L)^{1/2} \left(1+\frac{x^{(1-\theta )/2}}{(\alpha \omega )^{1/2}} \right) + \frac{\delta }{\alpha x^{\theta }} \ \pi _C(x_j,L) \\&\quad + \frac{\alpha x^{\theta }}{T_1} \log (\alpha x^{\theta }) \pi _C(x_j,L) + \frac{1}{U_{\pm }} \int _{T_0}^{T_1} |L(1/2+it)| |F(-\theta /2-i \theta t)| ~dt\) 

In order to estimate the integral in the error term above, we use dyadic division, and bound the following integrals
\(\frac{1}{2\pi i} \int _{1/2+iT^{\prime }}^{1/2+i2T^{\prime }} L(s) H(s)F(-s)~ds \ll \frac{1}{U}_{\pm } \int _{T^{\prime }}^{2T^{\prime }} |L(1/2+it)| |F(-\theta /2-i \theta t)| ~dt\) 
for \(T_0 \le T^{\prime } \le T_1/2\)  using Cauchy-Schwarz inequality. This is achieved by obtaining the following uniform bound for \(T^{\prime } \le \tau \le 2T^{\prime }\)  and \([x_j, x_{j+1}]\subseteq [x,2x] \)  established in Proposition REF  for \(K=\mathbb {Q}\)
\(F(-\theta /2-i \theta \tau ) &\ll x^{\theta /2}\left(\log d_L+b+\frac{bx\log x}{T^{\prime }} \right) \\& \quad + \frac{|C|}{|G|}x^{(1+\theta )/2}\bigg (\log d_L +n_L\log T^{\prime }\bigg ) \left(\frac{\log T^{\prime }}{\log x} +\frac{x^{1/2}}{T^{\prime }} \right);\) 

and the mean value estimate below given by Lemma  REF
\(\int _{T^{\prime }}^{2T^{\prime }}|L(1/2+it)|^2~dt \ll \alpha {T^{\prime }} + \alpha ^2 x^{\theta } (\log \alpha x^{\theta }).\) 
This gives us
\(\nonumber &\int _{T^{\prime }}^{2T^{\prime }} |L(1/2+it)| |F(-\theta /2 - i \theta t)| ~dt \\&\ll \left( \alpha {T^{\prime }}^2 + \alpha ^2 x^{\theta } T^{\prime }(\log \alpha x^{\theta }) \right)^{1/2}\bigg \lbrace x^{\theta /2}\left(\log d_L +b +b \frac{x\log x}{T^{\prime }} \right) \\&\quad +\frac{|C|}{|G|}x^{(1+\theta )/2}\bigg (\log d_L +n_L\log T^{\prime }\bigg ) \left(\frac{\log T^{\prime }}{\log x} +\frac{x^{1/2}}{T^{\prime }} \right)\bigg \rbrace & {\vspace{28.45274pt}}\\&\ll x^{\theta /2} \left(\log d_L + b \right) \left(\alpha {T^{\prime }}^2 + \alpha ^2 T^{\prime } x^\theta \log (\alpha x^{\theta })\right)^{1/2}\\&\quad + b x^{1+\theta /2} \log x \left(\alpha + \alpha ^2 {T^{\prime }}^{-1} x^\theta \log (\alpha x^{\theta }) \right)^{1/2}\\&\quad +\frac{|C|}{|G|} x^{(1+\theta )/2} \left(\log d_L + n_L\log T^{\prime }\right)\\&\quad \left(\left(\alpha {T^{\prime }}^2 + \alpha ^2 T^{\prime } x^{\theta } \log (\alpha x^{\theta })\right)^{1/2} \frac{\log T^{\prime } }{\log x}+ \left(\alpha + \alpha ^2 {T^{\prime }}^{-1} x^{\theta } \log (\alpha x^{\theta }) \right)^{1/2} x^{1/2}\right).\) 

Recall that \(U_{\pm }\ll \dfrac{\alpha x^{\theta }}{\delta }\) , \(\alpha x^{\theta } \le T^{\prime } < T_1\) , and \( \alpha x^{\theta }\log (\alpha x^{\theta })\ll {\delta } T_1\) . Therefore,
\(& \nonumber \frac{1}{U_{\pm }} \int _{T^{\prime }}^{2T^{\prime }} |L(1/2+it)| |F(-\theta /2 - i \theta t)| ~dt \\&\ll \frac{\delta }{(\alpha x^\theta )^{1/2}}\ \bigg \lbrace \frac{|C|}{|G|} x^{1/2} (\log d_L + n_L\log T_1) \left(T_1 \frac{\log T_1}{\log x}+x^{1/2}\right) \\&\quad + T_1 \left(\log d_L + b\right) + b x \log x \bigg \rbrace .\) 

We now set
\(T_1 &=\frac{\alpha ^{3/4}}{(n_L\delta \omega )^{1/2}\log x}x^{(1 +3\theta )/4}.\) 

Note that \(\log T_1 \ll \log x\)  since we have assumed \(\displaystyle {\alpha ^{1/4}(\omega n_L/ \delta )^{1/2}(\log x)^2 \ll x^{(1-\theta )/4}}.\)  Thus, using \(\log d_L \ll n_L\log n_L\)  (which follows from the discussion in Serre[2]}), we conclude
\(\nonumber \frac{1}{U_{\pm }} \int _{T^{\prime }}^{2T^{\prime }} &|L(1/2+it)| |F(-\theta /2 - i \theta t)| ~dt \ll (\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{(1 +\theta )/4}\\& \quad + \frac{|C|}{|G|}\left((\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{3/4+\theta /4} + \frac{\delta n_L}{\alpha ^{1/2}}x^{1-\theta /2}\log x\right).\) 

Given the value of \(T_1\) , since we use dyadic division of the interval \([T_0, T_1]\) , the number of integrals that we need to add is \(\operatorname{O}(\log x)\) . With this, inserting the above estimate in (REF ) gives us
\(&\nonumber \sum _{\begin{array}{c}x_j<p\le x_{j+1}\\ \sigma _p \in C\end{array}}\left(\left[\alpha p^\theta \ - \delta _1 \right] - \left[\alpha p^\theta \left(1-\frac{1}{U_{\pm }} \right) -\delta _1 \right] - \frac{\alpha p^{\theta }}{U_{\pm }}\right) \\& \ll \delta \pi _C(x_j,L)^{1/2} \left(1+\frac{x^{(1-\theta )/2}}{(\alpha \omega )^{1/2}} \right) + \frac{\delta }{\alpha x^{\theta }} \ \pi _C(x_j,L) \\&  \quad + \frac{|C|}{|G|}\log x \left( (\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{(3+\theta )/4} + \frac{\delta n_L}{\alpha ^{1/2}}x^{1-\theta /2}\log x\right)\\& \nonumber  \quad + (\delta n_L/\omega )^{1/2}\alpha ^{1/4}x^{(1+\theta )/4}\log x.\) 

Note that the error terms in (REF ) can be absorbed into (). Invoking (REF ), this completes the proof of Theorem REF .
